Write the standard form of the equation of the parabola whose directrix x = -1 and whose focus is at (5, -2)

a. (y + 2)^2 = 12 (x + 2)
b. y - 2 = 12 (x + 2)^2
c. x +2 + 1/12 (y + 2)^2
d. x - 2 = 1/12 (y + 2)^2

The directrix is a vertical line. Therefore, the parabola is of the form (y-k)^2=4p(x-h)

Consider the points (5,-2) and (-1,-2) = distance between the focus and the directrix = 2p
2p = sqrt[ (5+1)^2 + (-1-(-2))^2] = 6
2p=6
p=3

(y-k)^2 = 4(3)(x-h)
(y-k)^2 = 12(x-h)

The distance between the focus (5,-2) and the vertex (h,-2) is p
3 = sqrt[ (5-h)^2 +0^2 ] = (5-h)
h = 5-3 = 2

(y+2)^2 = 12( x-2)
(x-2) = (1/12) (y+2)^2 is the equation of the parabola

If the directrix is x = -1 and the x coordinate of the focus is 5 then the parabola opens to the right. The x coordinate of the vertex will be half way between -1 and 5, 2, and the y coordinate will be the same as the focus, -2. Therefore, the vertex is (2, -2).
Using the vertex form:
x = a(y + 2)² + 2
We could move the 2 to the left side and compute a but the only equation that fits the form is D.

vertex halfway between focus and directrix gives you (2,-2), and opens left, so
a(x - 2) = (y + 2)²
and a is 4 times the focal length, so 4(3) = 12, which makes the equation

12(x - 2) = (y + 2)² ∙ ∙ ∙ which matches d